## Calculus: Early Transcendentals 8th Edition

$f'(x)=\dfrac{x^{4}e^{x}+2xe^{2x}}{(x^{2}+e^{x})^{2}}$
$f(x)=\dfrac{x^{2}e^{x}}{x^{2}+e^{x}}$ Differentiate using the quotient rule: $f'(x)=\dfrac{(x^{2}+e^{x})(x^{2}e^{x})'-(x^{2}e^{x})(x^{2}+e^{x})'}{(x^{2}+e^{x})^{2}}=...$ Apply the product rule to evaluate $(x^{2}e^{x})'$: $...=\dfrac{(x^{2}+e^{x})[(x^{2})(e^{x})'+(e^{x})(x^{2})']-(x^{2}e^{x})(2x+e^{x})}{(x^{2}+e^{x})^{2}}=...$ $\dfrac{(x^{2}+e^{x})(x^{2}e^{x}+2xe^{x})-(x^{2}e^{x})(2x+e^{x})}{(x^{2}+e^{x})^{2}}=...$ Simplify: $...=\dfrac{x^{4}e^{x}+2x^{3}e^{x}+x^{2}e^{2x}+2xe^{2x}-2x^{3}e^{x}-x^{2}e^{2x}}{(x^{2}+e^{x})^{2}}=...$ $...=\dfrac{x^{4}e^{x}+2xe^{2x}}{(x^{2}+e^{x})^{2}}$